Question

Help with calc? :P

Answer 1

Find the centroid bounded by the curves: \(x+y=2\) and \(x=y^2\)

Answer 2

How far can you get on your own?

Answer 3

I got through the problem, but ended with a totally wrong answer xD

I think it's because I started with the wrong equations and such..

Answer 4

http://tutorial.math.lamar.edu/Classes/CalcII/CenterOfMass.aspx

Does this help?

btw, can someone explain to me the equation of the moments there?

Answer 5

I think it's the dA that's wrong

\[\Large (\frac{\int(x)((2-x)-\sqrt{x})dx}{\int((2-x)- \sqrt{x} )dx}, \frac{\frac{1}{2}\int((2-x)+sqrt{x})(2-x-\sqrt{x}) dx}{\int((2-x)-\sqrt{x})dx})\]

Answer 6

\[(\overline{x} , \overline{y}) = \left(\frac{1}{A} \iint x dA ,~~ \frac{1}{A} \iint y dA\right)\]

Answer 7

start by finding the area bounded by the curves : \(A\)

Answer 8

We're not into double integrals so idk what that means >.<

Answer 9

oh sorry, here is the single variable equivalent :

\[(\overline{x} , \overline{y}) = \left(\frac{1}{A} \int x(f-g) ~dx ,~~ \frac{1}{2A} \int f^2 - g^2 ~dx\right)\]

Answer 10

are you using the same/similar formula ?

Answer 11

This is the formula we were told work with..
\[\Large (\overline{x}_{c}, \overline{y}_{c})=(\frac{\int x dA}{\int dA}, \frac{\int y dA}{\int dA})\]

Answer 12

dA has to be used with double integral
dx has to be used with single integral

Answer 13

in your formula, both notations are mixed

Answer 14

your formula should look like below :

\[\large (\overline{x}_{c}, \overline{y}_{c})=\left(\frac{\iint x dA}{\iint dA}, \frac{\iint y dA}{\iint dA}\right)\]

Answer 15

That formula is easily derived. It is just a weighted average.

Answer 16

You have intersections at \((1,1)\) and at \((4,-2)\).

Answer 17

Well idk gane, I'm just working with what was given to me ._.

And I drew it out and everything wio, I think I just got the wrong equations..

Answer 18

Try doing without the equations.

Answer 19

Remember what we're really talking about here. Here is the formula you really want to know:
Given a function \(f(x)\) and weights \(w(x)\), the weighted average is given by: \[ \large
\frac{\int f(x)w(x)~dx}{\int w(x)~dx}
\]

Answer 20

Okay, so if you want the central \(x\) coordinate then, the weight will be \(x\) if we integrate with respect to \(x\) or \(x=f(y)\) if we integrate with respect to \(y\). The weight is just going to be the length of the slice for said \(x\) value.